Answer
$4$
Work Step by Step
$f$ is defined for all $x$ near $a$ with $x\lt a,$ as $f(x)=g(x).$
Then, $\displaystyle \lim_{x\rightarrow 3^{-}}f(x)=\lim_{x\rightarrow 3^{-}}g(x)=4,$
because the left-sided limit of g exists, and equals the limit of g(x) at 3.
$f$ is defined for all $x$ near $a$ with $x\gt a,$ as $f(x)=g(x).$
Then, $\displaystyle \lim_{x\rightarrow 3^{+}}f(x)=\lim_{x\rightarrow 3^{+}}g(x)=4,$
because the right-sided limit of g exists, and equals the limit of g(x) at 3.
We have now, at x=3, both the one-sided limits of f(x) exist and are equal to 4.
Then, the limit of f at x=3 exists and
$\displaystyle \lim_{x\rightarrow 3}f(x)=4$
